chemistry - exam practice help!!
posted by Anya on .
Given the table below predict the numerical value of the standard cell potential for the reaction:
2 Cr(s) + 3 Cu2+(aq) 2 Cr3+(aq) + 3 Cu(s)
Half Reaction E (volts)
(1) Cr3+ + 3 e- Cr E= -0.74
(2) Cr3+ + e- Cr2+ E=-0.41
(3) Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
(4) Cu+ + e- Cu E= 0.52
(5) Cu2+ + 2 e- Cu E= 0.34
(6) Cu2+ + e- Cu+ E= 0.16
how do i do this?
a ) -0.4
(all units are volts)
E(products) - E(reactants) but it didn't work, i didn't get any of the options. :S
First, you have no arrows. How can you possibly know which are products and which are reactants. I think you do yourself a disservice by not including an arrow.
For 2Cr + 3Cu^2+ ==> 3Cu + 2Cr^3+
Divide this into its two half cells.
Cr(s) ==> Cr^3+ + 3e
Cu^2+ + 2e ==> Cu(s)
Now look at the half cell voltages given to you. 1 is the one you want for Cr. That is
Cr3+ + 3 e- ==> Cr E= -0.74 which is just the reverse of your half cell. Therefore, the voltage for the reverse reaction is +0.74.
For the Cu, you want #5
Cu2+ + 2 e- ==> Cu E= 0.34
Eocell is Eox + Ered = 0.74+0.34 = ?